Question: Let $C$ be the line defined by $y = \dfrac{-x}{2} + 1$. We have a change of variables: $\begin{aligned} x &= X_1(u, v) = \dfrac{u}{3} + \dfrac{v}{2} \\ \\ y &= X_2(u, v) = \dfrac{u}{3} - \dfrac{v}{2} \end{aligned}$ What is $C$ under the change of variables? Choose 1 answer: Choose 1 answer: (Choice A) A $2u - v = 4$ (Choice B) B $u - 3v = 4$ (Choice C) C $2u - 3v = 4$ (Choice D) D $u- v = 4$
When applying a change of variables, we substitute the new definition for $x$ and $y$ into the original equation. The original equation: $y = \dfrac{-x}{2} + 1$ Let's substitute $X_1(u, v)$ for $x$ and $X_2(u, v)$ for $y$. $\begin{aligned} \dfrac{u}{3} - \dfrac{v}{2} &= \dfrac{-1}{2} \left( \dfrac{u}{3} + \dfrac{v}{2} \right) + 1 \\ \\ \dfrac{u}{3} - \dfrac{v}{2} &= \dfrac{-u}{6} - \dfrac{v}{4} + 1 \\ \\ 4u - 6v &= -2u - 3v + 12 \\ \\ 6u - 3v &= 12 \\ \\ 2u - v &= 4 \end{aligned}$ Therefore, under the change of variables, $C$ becomes: $2u - v = 4$